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3 years ago

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An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

biased in the normal active mode, with an emitter-base voltage of 1V. If the neutral base width is 100 nm, the emitter is 200 nm wide, and we have negligible base recombination, calculate the emitter current, emitter injection efficiency, and base transport factor

1 answer:

Darina [25.2K]3 years ago

8 0

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

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Suppose the Bookstore is processing an input file containing the titles of books in order to remove duplicates from their list.

mash [69]

Answer:

books = []

fp = open("bookTitles.txt")

for line in fp.readlines():

title = line.strip()

if title not in books:

books.append(title)

fp.close()

fout = open("noDuplicates.txt", "w")

for title in books:

print(tile, file=fout)

fout.close()

except FileNotFoundError:

print("Unable to open bookTitles.txt")

6 0

3 years ago

Write a program that asks the user for the name of a file. The program should display the number of words that the file contains

Flura [38]

Answer:

import java.io.*;

import java.util.Scanner;

public class CountWordsInFile {

public static void main(String[] args) throws IOException {

Scanner keyboard = new Scanner(System.in);

System.out.print("Enter file name: ");

String fileName = keyboard.next();

File file = new File(fileName);

try {

Scanner scan = new Scanner(file);

int count = 0;

while(scan.hasNext()) {

scan.next();

count += 1;

}

scan.close();

System.out.println("Number of words: "+count);

} catch (FileNotFoundException e) {

System.out.println("File " + file.getName() + " not present ");

System.exit(0);

}

}

}

3 0

3 years ago

What will the following segment of code output? score = 95; if (score > 95) cout << "Congratulations!\n"; cout <<

Anarel [89]

Answer:

That's a high score!

This is a test question!

Explanation:

The reason these two lines are printed and not the first one is simple. After the 'IF' condition has been stated, there is no use of parenthesis such as { and } to enclose the next lines. This means that only the first line after the 'IF' condition may be read or skipped depending on whether the condition (score>95) is met. Since the score is not larger than 95, and the 'IF' condition fails, the line 'Congratulations!' is not printed. The next two lines of the code are read as normal because they do not depend on the 'IF' condition.

5 0

3 years ago

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

3 years ago

A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut

denis23 [38]

Answer:

a) $T_m=1.787min$

b) $MRR=35259.7mm^3/min$

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Length $l=0.7m=>700mm$

Width $w=30mm$

Feed $F=0.25mm/tooth$

Depth $dp=3mm$

Diameter $d=75mm$

Number of cutting teeth $n=8$

Rotation speed $N=200rpm$

Generally the equation for Approach is mathematically given by

$x=\sqrt{Dd-d^2}$

$X=\sqrt{75*3-3^2}$

$X=14.69mm$

Therefore

Effective length is given as

$L_e=Approach +object Length$

$L_e=700+14.69$

$L_e=714.69mm$

a)

Generally the equation for Machine Time is mathematically given by

$T_m=\frac{L_e}{F_m}$

Where

$F_m=F*n*N$

$F_m=0.25*8*200$

$F_m=400$

Therefore

$T_m=\frac{714.69}{400}$

$T_m=1.787min$

b)

Generally the equation for Material Removal Rate. is mathematically given by

$MRR=\frac{L*B*d}{t_m}$

$MRR=\frac{700*30*3}{1.787}$

$MRR=35259.7mm^3/min$

3 0

3 years ago